3.7.60 \(\int (1+x) \sqrt {-1+x^2} \, dx\)

Optimal. Leaf size=44 \[ \frac {1}{3} \left (x^2-1\right )^{3/2}+\frac {1}{2} x \sqrt {x^2-1}-\frac {1}{2} \tanh ^{-1}\left (\frac {x}{\sqrt {x^2-1}}\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {641, 195, 217, 206} \begin {gather*} \frac {1}{3} \left (x^2-1\right )^{3/2}+\frac {1}{2} x \sqrt {x^2-1}-\frac {1}{2} \tanh ^{-1}\left (\frac {x}{\sqrt {x^2-1}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + x)*Sqrt[-1 + x^2],x]

[Out]

(x*Sqrt[-1 + x^2])/2 + (-1 + x^2)^(3/2)/3 - ArcTanh[x/Sqrt[-1 + x^2]]/2

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 641

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*(a + c*x^2)^(p + 1))/(2*c*(p + 1)),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

Rubi steps

\begin {align*} \int (1+x) \sqrt {-1+x^2} \, dx &=\frac {1}{3} \left (-1+x^2\right )^{3/2}+\int \sqrt {-1+x^2} \, dx\\ &=\frac {1}{2} x \sqrt {-1+x^2}+\frac {1}{3} \left (-1+x^2\right )^{3/2}-\frac {1}{2} \int \frac {1}{\sqrt {-1+x^2}} \, dx\\ &=\frac {1}{2} x \sqrt {-1+x^2}+\frac {1}{3} \left (-1+x^2\right )^{3/2}-\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x}{\sqrt {-1+x^2}}\right )\\ &=\frac {1}{2} x \sqrt {-1+x^2}+\frac {1}{3} \left (-1+x^2\right )^{3/2}-\frac {1}{2} \tanh ^{-1}\left (\frac {x}{\sqrt {-1+x^2}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 49, normalized size = 1.11 \begin {gather*} \frac {\left (x^2-1\right ) \left (\sqrt {1-x^2} \left (2 x^2+3 x-2\right )+3 \sin ^{-1}(x)\right )}{6 \sqrt {-\left (x^2-1\right )^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + x)*Sqrt[-1 + x^2],x]

[Out]

((-1 + x^2)*(Sqrt[1 - x^2]*(-2 + 3*x + 2*x^2) + 3*ArcSin[x]))/(6*Sqrt[-(-1 + x^2)^2])

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IntegrateAlgebraic [A]  time = 0.13, size = 42, normalized size = 0.95 \begin {gather*} \frac {1}{6} \sqrt {x^2-1} \left (2 x^2+3 x-2\right )-\tanh ^{-1}\left (\frac {\sqrt {x^2-1}}{x-1}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(1 + x)*Sqrt[-1 + x^2],x]

[Out]

(Sqrt[-1 + x^2]*(-2 + 3*x + 2*x^2))/6 - ArcTanh[Sqrt[-1 + x^2]/(-1 + x)]

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fricas [A]  time = 0.41, size = 34, normalized size = 0.77 \begin {gather*} \frac {1}{6} \, {\left (2 \, x^{2} + 3 \, x - 2\right )} \sqrt {x^{2} - 1} + \frac {1}{2} \, \log \left (-x + \sqrt {x^{2} - 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)*(x^2-1)^(1/2),x, algorithm="fricas")

[Out]

1/6*(2*x^2 + 3*x - 2)*sqrt(x^2 - 1) + 1/2*log(-x + sqrt(x^2 - 1))

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giac [A]  time = 0.17, size = 34, normalized size = 0.77 \begin {gather*} \frac {1}{6} \, {\left ({\left (2 \, x + 3\right )} x - 2\right )} \sqrt {x^{2} - 1} + \frac {1}{2} \, \log \left ({\left | -x + \sqrt {x^{2} - 1} \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)*(x^2-1)^(1/2),x, algorithm="giac")

[Out]

1/6*((2*x + 3)*x - 2)*sqrt(x^2 - 1) + 1/2*log(abs(-x + sqrt(x^2 - 1)))

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maple [A]  time = 0.04, size = 33, normalized size = 0.75 \begin {gather*} \frac {\sqrt {x^{2}-1}\, x}{2}-\frac {\ln \left (x +\sqrt {x^{2}-1}\right )}{2}+\frac {\left (x^{2}-1\right )^{\frac {3}{2}}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x+1)*(x^2-1)^(1/2),x)

[Out]

1/3*(x^2-1)^(3/2)+1/2*x*(x^2-1)^(1/2)-1/2*ln(x+(x^2-1)^(1/2))

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maxima [A]  time = 1.33, size = 36, normalized size = 0.82 \begin {gather*} \frac {1}{3} \, {\left (x^{2} - 1\right )}^{\frac {3}{2}} + \frac {1}{2} \, \sqrt {x^{2} - 1} x - \frac {1}{2} \, \log \left (2 \, x + 2 \, \sqrt {x^{2} - 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)*(x^2-1)^(1/2),x, algorithm="maxima")

[Out]

1/3*(x^2 - 1)^(3/2) + 1/2*sqrt(x^2 - 1)*x - 1/2*log(2*x + 2*sqrt(x^2 - 1))

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mupad [B]  time = 0.56, size = 32, normalized size = 0.73 \begin {gather*} \frac {x\,\sqrt {x^2-1}}{2}-\frac {\ln \left (x+\sqrt {x^2-1}\right )}{2}+\frac {{\left (x^2-1\right )}^{3/2}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2 - 1)^(1/2)*(x + 1),x)

[Out]

(x*(x^2 - 1)^(1/2))/2 - log(x + (x^2 - 1)^(1/2))/2 + (x^2 - 1)^(3/2)/3

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sympy [A]  time = 0.21, size = 39, normalized size = 0.89 \begin {gather*} \frac {x^{2} \sqrt {x^{2} - 1}}{3} + \frac {x \sqrt {x^{2} - 1}}{2} - \frac {\sqrt {x^{2} - 1}}{3} - \frac {\operatorname {acosh}{\relax (x )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+x)*(x**2-1)**(1/2),x)

[Out]

x**2*sqrt(x**2 - 1)/3 + x*sqrt(x**2 - 1)/2 - sqrt(x**2 - 1)/3 - acosh(x)/2

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